Question

Consider a solid copper sphere of radius R and a hollow copper sphere of outer radius R, both heated to the same temperature. These spheres are then allowed to cool in the same surroundings. State which of the following statements is correct and give reasons.
A) The solid sphere cools faster than the hollow sphere.
B) The hollow sphere cools faster than the solid sphere.
C) Both the spheres cool equally fast.
D) It cannot be decided.

Answer 1

Hint:The rate of change in temperature of a body that is allowed to cool depends on the amount of heat given out to the environment, the specific heat capacity of the material of the body and mass of the body. A solid sphere is more massive than a hollow sphere.

Formula used:
-The rate of change in temperature $T$ of an object is given by, $\dfrac{{dT}}{{dt}} = \dfrac{1}{{ms}}\left( {\dfrac{{dQ}}{{dt}}} \right)$ where $m$ is the mass of the object, $s$ is the specific heat capacity and $\dfrac{{dQ}}{{dt}}$ is the rate of change of heat.

Complete step by step answer.
Step 1: List the data given in the question.
Given, the radius of the solid sphere $R$ is equal to the outer radius of the hollow sphere. This implies that both the spheres have the same surface area.
These spheres were heated to the same temperature ( ${T_s} = {T_h}$ ) and then allowed to cool in the same environment.
Since both the spheres are made of copper they have the same specific heat capacity as well.
But the masses of the solid sphere and hollow sphere differ. Let ${m_s}$ be the mass of the solid sphere and ${m_h}$ be the mass of the hollow sphere. Then ${m_s} > {m_h}$ .
Step 2: Express the relation for the rate of change in temperature for both the spheres.
The relation for the rate of change in temperature is given by, $\dfrac{{dT}}{{dt}} = \dfrac{1}{{ms}}\left( {\dfrac{{dQ}}{{dt}}} \right)$ -------- (1)
where $m$ is the mass of the object, $s$ is the specific heat capacity and $\dfrac{{dQ}}{{dt}}$ is the rate of change of heat.
For the solid sphere made of copper, equation (1) will be $\dfrac{{d{T_s}}}{{dt}} = \dfrac{1}{{{m_s}s}}\left( {\dfrac{{dQ}}{{dt}}} \right)$ ------- (2)
For the hollow sphere made of copper, equation (2) will be $\dfrac{{d{T_h}}}{{dt}} = \dfrac{1}{{{m_h}s}}\left( {\dfrac{{dQ}}{{dt}}} \right)$ -------- (3)
Step 3: Based on the above equations determine which of the two spheres cool faster.
We know that the mass of the solid sphere is greater than the hollow sphere, i.e., ${m_s} > {m_h}$ .
Sine, both the spheres are allowed to cool in the same surrounding the rate of change in heat $\dfrac{{dQ}}{{dt}}$ must be the same for both of the spheres.
Therefore, equation (2) implies that the rate of change in temperature of the solid sphere is inversely proportional to its mass, i.e., $\dfrac{{d{T_s}}}{{dt}} \propto \dfrac{1}{{{m_s}}}$.
Similarly, equation (3) implies that the rate of change in temperature of the hollow sphere is inversely proportional to its mass, i.e., $\dfrac{{d{T_h}}}{{dt}} \propto \dfrac{1}{{{m_h}}}$.
Then as ${m_s} > {m_h}$ , $\dfrac{{d{T_s}}}{{dt}} < \dfrac{{d{T_h}}}{{dt}}$

Therefore, the hollow sphere cools faster than the solid sphere.

Note: A body that cools faster than another body will have a greater rate of change in its temperature i.e., the change in the temperature of that body will be more than the other one for the same amount of time.